What am I doing wrong with this proof?

**terrorsquid** · August 16th 2011, 09:35 AM

So, I'm trying to use induction to prove:

$\sum_{k=0}^n\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n},~~$ $~~for~all~n \in \mathbb{N}$

Base case is true.

Assume:

$\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n}$

Test:

$\left(\frac{-1}{3}\right)^{k+1} = \frac{3}{4} + \frac{1}{4}(-1)^{(n+1)}3^{-(n+1)}$

$\left(\frac{-1}{3}\right)^k\left(\frac{-1}{3}\right) = \frac{3}{4} + \frac{1}{4}(-1)^{n}(-1)3^{-n}3^{-1}$

$\left(\frac{3}{4} + \frac{1}{4}(-1)^n3^{-n}\right)\left(\frac{-1}{3}\right) = \frac{3}{4} -\frac{1}{12}(-1)^{n}3^{-n}$

This next step is what seems to cause the problems. If the 3/4 on the LHS was not in brackets, everything would be fine and become = when I multiply the other term by -1/3; however, by multiplying the 3/4 by -1/3 it makes them not equal:

$-\frac{1}{4} - \frac{1}{12}(-1)^n3^{-n} = \frac{3}{4} -\frac{1}{12}(-1)^{n}3^{-n}$

What am I doing wrong?

**topspin1617** · August 16th 2011, 02:45 PM

What happened to all of the summation signs??

You're trying to prove something about the entire summation $\sum_{k=0}^n\left$\frac{-1}{3}\right$^k$ , not the individual term $\left$\frac{-1}{3}\right$^k$ .

**Plato** · August 16th 2011, 03:08 PM

Originally Posted by terrorsquid

So, I'm trying to use induction to prove:
$\sum_{k=0}^n\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n},~~$ $~~for~all~n \in \mathbb{N}$

Without induction.
Let $S_n = \sum\limits_{k = 0}^n {r^k }$

$\begin{align*} S_n&= 1+r+r^2+\cdots+r^n \\ rS_n &=r+r^2+r^3\cdots+r^{n+1}\\(1-r)S_n &=1-r^{n+1}\\ S_n &= \frac{1-r^{n+1}}{1-r} \end{align*}$

**terrorsquid** · August 16th 2011, 07:15 PM

Originally Posted by topspin1617

What happened to all of the summation signs??

It was a magic sum sign

haha, stayed up too late last night I think.

Thanks.

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